What is the integral of (x^4cos(x))/8 ?

The integral of (x^4cos(x))/8 is 1/8 (x^4sin(x)-4(-x^3cos(x)+3(x^2sin(x)-2(-xcos(x)+sin(x)))))+C

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integral of (x^4cos(x))/8

\int \frac{x ^{4} cos ( x )}{8} d x

\frac{1}{8} (x^{4} sin (x) - 4 (- x^{3} cos (x) + 3 (x^{2} sin (x) - 2 (- x cos (x) + sin (x))))) + C

Solution steps

\int \frac{x ^{4} cos ( x )}{8} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int x^{4} cos (x) d x

Apply Integration By Parts

: x^{4} sin (x) - \int 4 x^{3} sin (x) d x

= \frac{1}{8} (x^{4} sin (x) - \int 4 x^{3} sin (x) d x)

\int 4 x^{3} sin (x) d x = 4 (- x^{3} cos (x) + 3 (x^{2} sin (x) - 2 (- x cos (x) + sin (x))))

= \frac{1}{8} (x^{4} sin (x) - 4 (- x^{3} cos (x) + 3 (x^{2} sin (x) - 2 (- x cos (x) + sin (x)))))

Add a constant to the solution

= \frac{1}{8} (x^{4} sin (x) - 4 (- x^{3} cos (x) + 3 (x^{2} sin (x) - 2 (- x cos (x) + sin (x))))) + C

\int_{0}^{2} (4 x + 5) e^{- x} d x

\frac{d}{d x} (\frac{2 x ^{2} + 2 x + 4}{x})

y^{^{'}} = \frac{( x y + 2 y )}{( x + 3 )}

\int (4 x) d x

\frac{d y}{d t} = \frac{2 t}{1 + t ^{2}} y + \frac{2}{1 + t ^{2}}

What is the integral of (x^4cos(x))/8 ?
The integral of (x^4cos(x))/8 is 1/8 (x^4sin(x)-4(-x^3cos(x)+3(x^2sin(x)-2(-xcos(x)+sin(x)))))+C